AMC 10 · 2008 · #24
Grade 11 geometry-2dPick an answer.
The picture has exactly one degree of freedom, the length CA, so tool #4 (Introduce a Variable) turns the whole family of triangles into a single number x = CA ＞ 0 and tan∠ BAD into a function of x. Then tool #14 (Extreme Principle) is the actual question: maximize that function. The obstacle is that ∠ BAD sits inside triangle ABD, whose sides AB and AD both depend on x in ugly ways. Tool #15 (Organize Information in More Ways) removes the obstacle: instead of computing ∠ BAD itself, compute the two products AB · ADsin∠ BAD and AB · ADcos∠ BAD separately. The first is twice the area of ABD, which the midpoint makes trivial; the second is what the Law of Cosines hands over. Dividing them cancels the unknown product AB · AD and leaves tan∠ BAD as a clean rational function of x — with no case split over acute versus obtuse, which is where the more common "subtract two angles" route gets delicate. Tool #1 (Draw a Diagram) keeps the configuration honest: D between B and C is what guarantees ray AD lies inside angle BAC. One warning is built into the plan: an inequality only produces a ceiling, so the last step must exhibit a real triangle that touches it.
Name the one free length
One free length carries the whole configuration.
One number decides the entire picture, so the geometry question is really a one-variable question.
9.A-CED.A.2Introduce A VariableGet the sine part from area
An area gives the sine part directly.
A median splits a triangle into two equal areas, so the awkward half is as easy as the whole.
11.G-SRT.D.9Draw A DiagramGet the cosine part from the Law of Cosines
The law of cosines gives the cosine part.
The Law of Cosines is exactly the tool that trades an unknown angle for the three sides around it.
11.G-SRT.D.10Organize Information In More WaysDivide and read off the tangent
Dividing reads off the tangent as one fraction.
Tangent is sine over cosine, so two products of the same two sides divide into the angle itself.
10.G-SRT.C.6Organize Information In More WaysFlip it into a sum to minimize
Flipping it turns maximizing into minimizing a sum.
A sum of a number and its reciprocal multiple is smallest when the two pieces are equal, and a squared difference is the proof.
A sum of a quantity and a fixed multiple of its reciprocal is smallest exactly when the two pieces are equal.
▸ Why?
With the product of the two pieces fixed, they balance best when they are equal, and pulling them apart only costs.
▸ Why?
A squared difference is never negative, so the gap between the sum and that balanced value can only be positive.
Show the ceiling is actually reached
The bound is actually reached, so the answer is √(3)/(4√(2)-3), choice (D).
A bound is only an answer once a real configuration sits exactly on it.
11.N-RN.A.2Extreme PrincipleWhen an angle sits between two sides you do not know, do not chase the angle — compute AB · ADsin∠ BAD from the area and AB · ADcos∠ BAD from the Law of Cosines, divide, and the unknown sides cancel and leave the tangent.
- Name the one free length
- Get the sine part from area
- Get the cosine part from the Law of Cosines
- Divide and read off the tangent
- Flip it into a sum to minimize
- Show the ceiling is actually reached